



Formulary
Definitions and basics
Trigonometric circle and angles o o o o
Trigonometric numbers of a real number t
Basic formulas o
Related values





The difference between the two values is an integer multiple of 2.pi supplementary values complementary values
Opposite values
Anti-supplementary values
The right-angled triangle
 Other properties in a right-angled triangle o o o
Area of a triangle
Sine rule
 Homogeneous expression in a, b and c
Cosine rule
Special values o pi/3 o pi/4 o pi/6
Solving Triangles o Case SSS o Case ASA or AAS o o o
Case SAS o Case SSA
 Trigonometric functions
The sine function
The cosine function
 o o o o
The tangent function
The cotangent function
Related functions and period
Period of a sum of two functions
Inverse Trigonometric Functions = Cyclometric functions o The arcsin function o The arccos function o o o o
The arctan function
The arccot function
No period
Transformations

 Sum formulas o cos(u - v) o cos(u + v) o o o o sin(u - v) sin(u + v) tan(u + v) tan(u - v)
 Doubling formulas o sin(2u) o cos(2u) o tan(2u)
 Carnot formulas

 t-Formulas or Half-Angle Formulas
 Simpson formulas
Factoring trigonometric forms


Period of the product of two related functions
 The general sine function
Trigonometric equations o Base equations
 cos(u) = cos(v)


 sin(u) = sin(v) tan(u) = tan(v) cot(u) = cot(v) o o o o
Reducing to base equations
Using an additional unknown
Using factoring
The equation a.sin(u)+b.cos(u) = c


First Method
Second Method
Homogeneous equations
 o o Other equations
Trigonometric inequalities o Conventions o Examples
 Cyclometric equations
 Calculations with cyclometric functions
 A Formulary
 Solved problems about trigonometry
 Other trigonometry tutorials
Click here to find basic formulas.

Choose an x-axis and a y-axis (orthonormal) and let O be the origin.
A circle of radius one centered at O is called 'the' trigonometric circle or 'the' unit circle.
Turning counterclockwise is the positive orientation in trigonometry.
Angles are measured starting from the x-axis.
The units used to measure an angle are 'degree' and 'radian'.
A right angle is an angle whose measure is exactly 90 degrees or pi/2 radians.
In this theory we use mainly radians.
Each real number t corresponds to exactly one angle, and to exactly one point P on the unit circle.
We call that point the 'image point' of t.
Examples:
 pi/6 corresponds to the angle t and to point P on the circle.
 -pi/2 corresponds to the angle u and to point Q on the circle.
The real number t corresponds to exactly one point P on the unit circle.
 The x-coordinate of P is called the cosine of t. We write cos(t).


The y-coordinate of P is called the sine of t. We write sin(t).


The number sin(t)/cos(t) is called the tangent of t. We write tan(t).
 The number cos(t)/sin(t) is called the cotangent of t. We write cot(t).
The number 1/cos(t) is called the secant of t. We write sec(t)
The number 1/sin(t) is called the cosecant of t. We write csc(t) or cosec(t)
The line with equation sin(t).x - cos(t).y = 0 contains the origin and point P(cos(t),sin(t)). So this line is OP.
On this line we take the intersection point S(1,?) with the line x = 1.

It is easy to see that ? = tan(t).
So tan(t) is the y-coordinate of the point S.
In an analogous manner we find that cotan(t) is the x-coordinate of the intersection point S' of the line OP with the line y = 1.
With t radians corresponds exactly one point P(cos(t),sin(t)) on the unit circle. The square of the distance [OP] = 1. Calculating |OP|
2
, using the coordinates of P, we find for each t : cos 2 (t) + sin 2 (t) = 1
sin 2 (t)
1 + tan 2 (t) = 1 + ----------
cos 2 (t)
cos 2 (t)+sin 2 (t)
= -----------------
cos 2 (t)
1
= ----------- = sec 2 (t)
cos 2 (t)
In the same way :
1 + cotan 2 (t) = 1/ sin 2 (t) = csc 2 (t)
cos 2 (t) + sin 2 (t) = 1
1 + tan 2 (t) = sec 2 (t)
1 + cot 2 (t) = csc 2 (t)
Usage examples:
sin 2 (t) = 1 - cos 2 (t)

cos 2 (4t) = 1 - sin 2 (4t)
1 + tan 2 (t/2) = sec 2 (t/2)
csc 2 (t 2 ) - cot 2 (t 2 ) = 1
Exercise:
If cos(t)=0.5 then sin 2 (t) = ...
If cos(t)=0.1 then tan 2 (t) = ...
If cot(t)=0.2 then sin 2 (t) = ...
The difference between the two values is an integer multiple of 2.pi
If the difference between t an t' is an integer multiple of 2.pi, the corresponding points on the unit circle coincide. So
If t - t' = 2.k.pi (k is an integer) then
sin(t) = sin(t')
cos(t) = cos(t')
tan(t) = tan(t')
cot(t) = cot(t') supplementary values t and t' are supplementary values <=> t+t' = pi.
With the help of a unit circle we see that the corresponding image points are symmetric relative to the y-axis. Hence, we have :
sin(t) = sin(pi - t)
cos(t) = -cos(pi - t)
tan(t) = -tan(pi - t)
cot(t) = -cot(pi - t)

Usage examples:
sin(t + pi/2) = sin(pi/2 - t)
tan(2t + 0.2) = - tan(pi -0.2 - 2t)
- tan(pi -t) = tan(t)
sin(pi-t) + cos(3pi-t) - sin(t+4pi) + cos(t)
= sin(t) + cos(pi-t) - sin(t) + cos(t)
= sin(t) - cos(t) - sin(t) + cos(t)
= 0 complementary values t and t' are complementary values <=> t+t' = pi/2.
The corresponding image points on a unit circle are symmetric relative to the line y = x . Hence, we have :
sin(t) = cos(pi/2 - t)
cos(t) = sin(pi/2 - t)
tan(t) = cot(pi/2 - t)
cot(t) = tan(pi/2 - t)
Usage examples:
tan(pi/4 +3t) = cot(pi/4 -3t)
cos(3pi/2 -t) = sin( t - pi) = sin(-t + 2pi) = sin(-t)
cot(3x - pi/2) = tan(-3x + pi ) = - tan(3x)
- cos(pi/2 - 2x) + sin(-2x - pi) - cos(3pi - 2x)
= - sin(2x) + sin(pi - 2x) - cos(pi - 2x)
= - sin(2x) + sin(2x) + cos(2x)
= cos(2x)

Opposite values t and t' are opposite values <=> t+t' = 0.
Now, the corresponding image points are symmetric relative to the x-axis. Hence, we have :
sin(t) = -sin(-t)
cos(t) = cos(-t)
tan(t) = -tan(-t)
cot(t) = -cot(-t)
Usage examples:
cos(-pi/2 + x) = cos(pi/2 - x) = sin (x)
sin(6x - pi) = - sin(pi - 6x) = - sin(6x)
cot(-x + 4pi) = cot(-x) = - cot(x)
Anti-supplementary values t and t' are anti-supplementary values <=> ( t-t' = pi or t'-t = pi )
The corresponding image points are symmetric relative to the origin O . Hence, we have :
sin(t) = -sin(t+pi)
cos(t) = -cos(t+pi)
tan(t) = tan(t+pi)
cot(t) = cot(t+pi)
Usage examples:
tan(5a + 3pi) = tan(5a + pi) = tan(5a)
cot(t/2 + pi/2) = cot(t/2 - pi/2) = - cot(pi/2 - t/2) = - tan(t/2)

sin(x + 3 pi) + sin(x) = -sin(x) + sin(x) = 0
Let the right angle of a triangle ABC be labelled A. The distances |AB|, |BC| and |CA| are usually denoted by c, a and b. Choose point B in a suitable way as center of a trigonometric circle (see figure).
Now sin(B),cos(B) and 1 are directly proportional with b, c and a.
sin(B) cos(B) 1
------ = ------ = ---
b c a
=> sin(B) = b/a cos(B) = c/a tan(B) = b/c and since the angles B and C are complementary angles
cos(C) = b/a sin(C) = c/a tan(C) = c/b
In each right-angled triangle ABC, with A as the right angle, we have
sin(B) = b/a cos(B) = c/a tan(B) = b/c
cos(C) = b/a sin(C) = c/a tan(C) = c/b
Other properties in a right-angled triangle
 The square of the hypotenuse is equal to the sum of the squares of the other two

sides. a 2 = b 2 + c 2
 The altitude to the hypotenuse of a right triangle is the mean proportional between the segments into which it divides the hypotenuse. h
2
= x.y
 Each leg of a right triangle is the mean proportional between the hypotenuse and the its orthogonal projection on the hypotenuse. c
2
= a.x and b
2
= a.y
Applications:
The tangent lines, in the points A and B of a circle with center O and radius r, intersect in point P. The chord AB and the line OP intersect in point S. Let a = |OP| and k =
|AB|.
Express k as a function of r and a.
In the right-angled triangle AOP : The leg of a right triangle is the mean proportional between the hypotenuse and the its orthogonal projection on the hypotenuse.
|OA| 2 = |OP| |OS|.
=> |OS| = r 2 /a
In the right-angled triangle OAS : The square of the hypotenuse is equal to the sum of the squares of the other two sides.
|OA| 2 = |OS| 2 + |AS| 2
=> r 2 = |OS| 2 + k 2 /4
=> r 2 = r 4 /a 2 + k 2 /4
=> ....

2r ___________
=> k = ---- \/ a 2 - r 2
a
Divide a given segment BC by the construction of a point D. The two parts BD and
DC must have a suitable length x and y such that the product x.y is equal to a given value m
2
. x+y=|BC| en m
2
= x.y m is the mean proportional between x and y
We know that the altitude to the hypotenuse BC of a right triangle ABC is the mean proportional between the segments into which it divides the hypotenuse.
We are looking for a right triangle ABC with base the hypotenuse BC and with m as the height.
The vertex A of the right-angled triangle is located on the circle with diameter BC.
Construction steps:
1.
Draw a semicircle with diameter BC
2.
Draw a parallel to BC at distance m from BC.
3.
This parallel gives us point A.
4.
Draw the altitude AD from A to BC.
The area of the triangle is a.h/2 .
But in triangle BAH, we have sin(B) = h/c .
Hence the area of the triangle is a.c.sin(B) /2.
Similarly, we have that the area of the triangle
= b.c.sin(A) /2 = a.b.sin(C) /2
The area of a triangle ABC =(1/2) a.c.sin(B) = (1/2) b.c.sin(A) = (1/2) a.b.sin(C)

You can also use Heron's formula to calculate the area of a triangle.
Let s = half the circumference of the triangle = (a +b + c)/2.
The area of a triangle ABC =
______________________________
V s (s - a) (s - b) (s - c)
Exercise:
A triangle has sides with length 5, 4 and 7.
Draw accurately the triangle.
Calculate the area of the triangle with the formula of Heron and check the result by measuring height of the triangle and calculating the area with this height. Now measure an angle of the triangle and calculate the area a third time. Conversely, you can calculate the angles of the triangle if you know the area of the triangle.
The area of a triangle ABC = a.c.sin(B)/2 = b.c.sin(A)/2 = a.b.sin(C)/2
=> a.c.sin(B) = b.c.sin(A) = a.b.sin(C) dividing through by a.b.c, we get
a b c
------ = ------ = ------
sin(A) sin(B) sin(C)
This formula is called the sine rule in a triangle ABC.
Let R be the radius of the circle with center O through the points A,B and C. Let B' be the second intersection point of BO with the circle. The angle B' in triangle BB'C is equal to, or supplementary with, A.
In the right-angled triangle BB'C we see that a = 2R sin(B') = 2R sin(A).
Thus, the fractions in the sinus rule are all equal to 2R.
In any triangle ABC we have

a b c
------ = ------ = ------ = 2R
sin(A) sin(B) sin(C)
Exercise:
A triangle has sides with length a = 5, b = 4 and c = 7.
Draw accurately the triangle. Calculate the area of the triangle with the formula of Heron.
Now calculate the angle A with the area formula (1/2).b.c.sin (A).
Now, use the angle A to find the radius R of the circumscribed circle.
Check the result on your sketch.
Homogeneous expression in a, b and c
Note:
A relation is called homogeneous in a, b and c if and only if this relation remains valid when we replace a, b and c by a multiple r.a, r.b and r.c (r not 0).
If an expression between the sides of a triangle is homogeneous in a, b and c, we obtain an equivalent expression by replacing a,b and c by sin(A), sin(B) and sin(C).
Example:
In a triangle
b.sin(A-C) = 3.c.cos(A+C)
<=>
sin(B).sin(A-C) = 3.sin(C).cos(A+C)
In any triangle ABC we have
a 2 = b 2 + c 2 - 2 b c cos(A)
b 2 = c 2 + a 2 - 2 c a cos(B)
c 2 = a 2 + b 2 - 2 a b cos(C)
Proof:
We'll prove that a
2
= b
2
+ c
2
- 2 b c cos(A)
If the angle A is a right angle, then the proof is obvious.
Now, suppose the angle A is an acute angle.

a 2 = h 2 + p 2 (*)
b 2 = h 2 + q 2
= h 2 + (c - p) 2
so,
h 2 = b 2 - (c - p) 2 (**)
From (*) and (**)
a 2 = b 2 - (c - p) 2 + p 2
= b 2 - (c 2 - 2 p c + p 2 ) + p 2
= b 2 - c 2 + 2 p c
= b 2 + c 2 + 2 p c - 2 c 2
= b 2 + c 2 + 2 c (p - c)
= b 2 + c 2 - 2 c (c - p)
= b 2 + c 2 - 2 c q
= b 2 + c 2 - 2 c b cos(A)
Now, suppose the angle A is an obtuse angle.
The proof proceeds in the same manner as above.
Draw a new picture and work this out as an exercise.
This cosine rule can also be proved using the dot product of vectors.
See Proof cosine rule
Let V be the image point corresponding with the angle pi/3 on the unit circle and let E the intersection point of that circle with the positive X-axis.
The triangle 'OVE' is equilateral. Hence cos(pi/3) = 1/2. sin 2 (pi/3) = sqrt( 1 - cos 2 (pi/3)) = sqrt(3)/2

So, sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2.
tan(pi/3) = sqrt(3)
Let V be the image point corresponding with the angle pi/4 on the unit circle. From this, it is obvious that cos(pi/4) = sin(pi/4) and tan(pi/4) = 1. cos 2 (pi/4)+sin 2 (pi/4) = 1 => 2cos 2 (pi/4) = 1 => cos (pi/4) = sqrt(1/2)
So, cos (pi/4) = sin(pi/4) = sqrt(1/2)
tan(pi/4) = 1
From properties for complementary angles we have: cos (pi/6) = sqrt(3)/2 and sin(pi/6) = 1/2. tan(pi/6) = 1/sqrt(3).
Given : The three sides.
Substitute all the sides in de Cosine Rule to compute the angles.
Example: a=4 b=5 c=7
The Cosine Rule gives
58 = 70 cos(A)
40 = 56 cos(B)
-8 = 40 cos(C)
A = 34.05° B = 44.41° C = 101.53°
Test : A + B + C = ...
Given : Two angles and a side.
Calculate the third angle and then the sides with the Sine Rule.
Example: a=4 A=34° B=45°

The third angle is C =101°
From the Sine Rule
4 sin(45°) b = -------------- = 5.06
sin(34°)
4 sin(101°) c = ------------- = 7.02
sin(34°)
Test : draw a sketch of the triangle
Given : Two sides and an included angle.
Use the Cosine Rule.
Example: b=5 c=7 A=34.05°
From the Cosine Rule a 2 = 25 + 49 - 70 cos(34.05°) => a = 4
The other two formulas of the cosine rule give
40 = 56 cos(B)
-8 = 40 cos(C)
B = 44.41° C = 101.53°
Test : A + B + C = ...
Given : Two sides and a non-included angle.
Draw a sketch. There are three cases.
1) no solutions
2) one solution
3) two solutions
1.
A=60° b=5 a=1
From a sketch we see that there are no solutions.
2.
A=60° b=5 a=7

From a sketch we see that there is one solution. We use the Sine Rule.
7 5 c
--------- = -------- = ---------
sin(60°) sin(B) sin(C)
So, sin(B)= 0.6186 and this gives us two supplementary solutions for B.
But from our sketch, we know what value to choose. B = 38.21°.
Then C = 180° - 38.21° - 60° = 81.79° and c = 8
3.
A=60° b=5 a=4.5
From a sketch we see that there are two solutions for B. We use the Sine Rule.
4.5 5 c
--------- = -------- = ---------
sin(60°) sin(B) sin(C)
So, sin(B)= 0.96225 and this gives us two supplementary solutions for B.
B = 74.2° of 105.8°
First choose B = 74.2° and first compute C and then c with the Sine Rule.
Then choose B = 105.8° and first compute C and then c with the Sine Rule.
Check the results using your sketch.
The function defined by : sin : R -> R : x -> sin(x) is called, the sine function.
The images are bounded in [-1,1] and the period is 2.pi .
We see that the range of the function is [-1,1].

The function defined by : cos : R -> R : x -> cos(x) is called, the cosine function.
The images are bounded in [-1,1] and the period is 2.pi .
The range of the function is [-1,1].
The function defined by : tan : R -> R : x -> tan(x) is called, the tangent function.
Now, the period is pi and the images are not defined in x = (pi/2) + k.pi
The range or image is R .
The function defined by : cot : R -> R : x -> cot(x) is called, the cotangent function.
The period is pi and that the images are not defined in x = k.pi
The range or image is R .

We can submit previous functions to all kinds of transformations. We obtain related functions. ( see Influence of a transformation on the graph of a function )
Example 1 y = sin(4x)
The graph of this function arises from the graph of sin(x) when we compress the graph of sin(x) towards the y-axis with a factor 4.
From this it follows that the period of sin(4x) is pi/2. The function y = sin(ax) has a period 2.pi/a if a > 0.
Similar rules apply to the other trigonometric functions. Thus the period of tan(x/3) is 3.pi.
Example 2 y = sin(x+5)
The graph of this function comes about by moving the graph of sin(x) five units to the left. The period does not change.
Example 3 y = tan(x)+5
The graph of this function is obtained by moving the graph of tan(x) five units upwards. The period does not change.
Example 4
We start with y = tan(x). We compress the graph towards the y-axis with a factor 3. The new function is y = tan(3x). We move the graph two units to the right. The new function is y = tan(3(x-2)) . Finally, we move the last graph two units downwards. We obtain y = tan(3x -6)-2.
The period is pi/3.
Generalization:

The period of A sin(a x + b ) is 2 pi/|a|
The period of A cos(a x + b ) is 2 pi/|a|
The period of A tan(a x + b ) is pi/|a|
The period of A cot(a x + b ) is pi/|a|
The period of A / sin(a x + b ) is 2 pi/|a|
The period of A / cos(a x + b ) is 2 pi/|a|
The period of A / tan(a x + b ) is pi/|a|
The period of A / cot(a x + b ) is pi/|a|
If
f(x) is a function with period = a
g(x) is a function with period = b
Then
f(x)+g(x) is a function with period = c
<=>
There are strictly positive and relatively prime integers m and n
such that c = m.a = n.b
Examples sin(2x) has pi as period and cos(3x) has 2pi/3 as period .
Now, c = 2.(pi) = 3.(2pi/3). So, 2 pi is a period of sin(2x) + cos(3x) sin(pi x) has 2 as period and tan(2 pi x/7) has 7/2 as period.
Now, c = 7.(2) = 4.(7/2). So, 14 is a period of sin(pi x) + tan(2 pi x/7) sin(sqrt(2) x) has pi.sqrt(2) as period and cos(2x) has pi as period .
There are no strictly positive integers m and n such that m.(pi.sqrt(2)) = n.(pi). So, sin(sqrt(2) x) + cos(2x) has NO period! sin(x) has 2pi as period and cos(pi x) has 2 as period.
There are no strictly positive integers m and n such that m.(2pi) = n.(2). So, sin(x) + cos(pi x) has NO period!
We restrict the domain of the sine function to [-pi/2 , pi/2].
Now this restriction is invertible because each image value in [-1,1] corresponds to exactly one original value in [-pi/2 , pi/2].
The inverse function of that restricted sine function is called the arcsine function.
We write arcsin(x) or asin(x).
The graph y = arcsin(x) is the mirror image of the restricted sine graph relative to the line y = x.
The domain is [-1,1] and the range is [-pi/2 , pi/2].

We restrict the domain of the cosine function to [0 , pi].
Now this restriction is invertible because each image value in [-1,1] corresponds to exactly one original value in [0 , pi].
The inverse function of that restricted cosine function is called the arccosine function.
We write arccos(x) or acos(x) .
The graph y = arccos(x) is the mirror image of the restricted cosine graph relative to the line y = x.
The domain is [-1,1] and the range is [0 , pi].
We restrict the domain of the tangent function to [-pi/2 , pi/2].
The inverse function of that restricted tangent function is called the arctangent function. We write arctan(x) or atan(x) . The graph y = arctan(x) is the mirror image of the restricted tangent graph relative to the line y = x.
The domain is R and the range is [-pi/2 , pi/2].

We restrict the domain of the cotangent function to [0 , pi].
The inverse function of that restricted cotangent function is called the arccotangent function.
We write arccot(x) or acot(x) .
The graph y = arccot(x) is the mirror image of the restricted cotangent graph relative to the line y
= x.
The domain is R and the range is [0 , pi].
The inverse trigonometric functions have no period!
As with the trigonometric functions, we can create related functions using simple transformations.
Example: y = 2.arcsin(x-1) comes about by moving the graph of arcsin(x) one unit to the right, and then by multiplying all the images by two. The domain is [0,2] and the range is [-pi,pi].
We prove this formula using the concept of dot product of two vectors. (See theory about vectors )
With u corresponds one point P(cos(u),sin(u)) on the unit circle
With v corresponds one point Q(cos(v),sin(v)) on the unit circle
The angle, corresponding with the arc QP on the circle, has a value u - v .
The dot product P .
Q = 1.1.cos(u-v) .
But using the coordinates we also have P .
Q = cos(u).cos(v)+sin(u).sin(v).
Hence, cos(u-v) = cos(u).cos(v)+sin(u).sin(v)

Example: cos(pi/3-2x) = cos(pi/3)cos(2x) + sin(pi/3)sin(2x) = 0.5 cos(2x) + 0.5 sqrt(3) sin(2x)
cos(u + v) = cos(u - (-v)) = cos(u).cos(-v)+sin(u).sin(-v) cos(u + v) = cos(u).cos(v)-sin(u).sin(v)
Example: cos(x + x/2) + cos(x - x/2) = cos(x)cos(x/2) + sin(x)sin(x/2) + cos(x)cos(x/2) - sin(x)sin(x/2)
= 2 cos(x)cos(x/2)
sin(u - v) = cos(pi/2-(u-v)) = cos( (pi/2-u) +v )
= cos(pi/2 - u).cos(v)-sin(pi/2 - u).sin(v) sin(u - v) = sin(u).cos(v)-cos(u).sin(v)
Example: sin(x - pi/4) = sin(x) cos(pi/4) - cos(x) sin(pi/4) = (sin(x)-cos(x))/sqrt(2)
sin(u + v) = cos(pi/2-(u+v)) = cos( (pi/2-u) -v )
= cos(pi/2 - u).cos(v)+sin(pi/2 - u).sin(v) sin(u + v) = sin(u).cos(v)+cos(u).sin(v)

sin(u + v) sin(u).cos(v)+cos(u).sin(v) tan(u+v) = ------------ = ---------------------------
cos(u + v) cos(u).cos(v)-sin(u).sin(v)
Dividing the dominator and denominator by cos(u).cos(v) we have
tan(u) + tan(v) tan(u+v) = -----------------
1 - tan(u).tan(v)
Example:
tan(u) + tan(pi/4) tan(u) + 1 1 + tan(u) tan(u+pi/4) = -------------------- = -------------- = -------------
1 - tan(u).tan(pi/4) 1 - tan(u) 1 - tan(u)
In the same way, we have
tan(u) - tan(v) tan(u-v) = -----------------
1 + tan(u).tan(v)
sin(2u) = sin(u + u) = sin(u).cos(u)+cos(u).sin(u) = 2sin(u).cos(u) sin(2u) = 2sin(u).cos(u)
Examples
sin(x) = 2 sin(x/2).cos(x/2)
sin(4x) = 2 sin(2x).cos(2x) = 4 sin(x) cos(x) cos(2x)
12 sin(8x) cos(8x) = 6 sin(16x)

cos(2u) = cos(u+u) = cos(u).cos(u)-sin(u).sin(u) = cos
2
(u) - sin
2
(u) cos(2u) = cos
2
(u) - sin
2
(u)
tan(u) + tan(u) 2 tan(u) tan(2u) = ------------------ = ---------------
1 - tan(u).tan(u) 1- tan(u)tan(u)
2 tan(u) tan(2u) = -----------
1- tan 2 (u)
Example:
1 cot(2x) = --------
tan(2x)
1 - tan 2 (x)
= -------------
2 tan(x)
1 + cos(2u) = 1+cos 2 (u)-sin 2 (u) = 2 cos 2 (u)
1 - cos(2u) = 1-cos 2 (u)+sin 2 (u) = 2 sin 2 (u)
1 + cos(2u) = 2 cos 2 (u)
1 - cos(2u) = 2 sin 2 (u)
Applications:
 From the carnot formulas, it follows that :
The period of cos(2u) = the period of cos
2
(u) = the period of sin
2
(u)
 Factor the expression 1 + 2 cos(x) + cos(2x)

 1 + 2 cos(x) + cos(2x)


 = 2 cos(x) + ( 1 + cos(2x))

 = 2 cos(x) + 2 cos 2 (x)

 = 2 cos(x) (1 + cos(x))

 = 2 cos(x) 2 cos 2 (x/2)

 = 4 cos(x) cos 2 (x/2)
Since 2 pi is the period of (1 + 2 cos(x) + cos(2x)), it follows that the period of cos(x) cos
2
(x/2) is 2pi.
 Find the period of tan
2
(4x)
The period of tan
2
(4x) is equal to the period of 1+tan
2
(4x).
The period of 1+tan
2
(4x) is equal to the period of 1/ cos
2
(4x).
The period of 1/ cos 2 (4x) is equal to the period of cos 2 (4x).
The period of cos
2
(4x) is equal to the period of 0.5(1+cos(8x)).
The period of 0.5(1+cos(8x)) is equal to the period of cos(8x).
And this period is pi/4.
 In triangle ABC the sides a, b, c are such that 3a = 7c en 3b = 8c.
Find tan
2
(A/2) without calculating A or A/2.
Solution:
About the three sides we know :
a b c
--- = --- = ---
7 8 3
Since similar triangles have the same angles, we can use a = 7 , b = 8 and c = 3 as sides of the triangle.
From the cosine rule we can write
b 2 + c 2 - a 2
cos(A) = ------------------ = 1/2
2 b c
Now we use the Carnot formulas
1 - cos(A) 2 sin 2 (A/2)
---------- = -------------- = tan 2 (A/2) = 1/3
1 + cos(A) 2 cos 2 (A/2)

 Find the exact value of cos(pi/12)
We know : 1 + cos(2u) = 2 cos
2
(u)
Now take 2u = pi/6 radians , then u = pi/12 radians.
Now the exact value of cos(pi/6) is sqrt(3)/2. We can use the Carnot formula to calculate cos(pi/12).
cos 2 (pi/12) = (1 + cos(pi/6))/2
= (1 + sqrt(3)/2)/2
= (2 + sqrt(3))/4
So, cos(pi/12) = (1/2). sqrt(2 + sqrt(3))
From the Carnot formulas we have cos(2u) = 2 cos 2 (u) -1
2
= ------------ - 1
1 + tan 2 (u)
1 - tan 2 (u)
= -------------
1 + tan 2 (u)
We know:
2 tan(u) tan(2u)= -------------
1 - tan 2 (u)
Hence,
2 tan(u) sin(2u) = -----------
1 + tan 2 (u)
Let t = tan(u) , then
1 - t 2 cos(2u) = ---------

1 + t 2 or
1 - tan 2 (u) cos(2u) = -------------
1 + tan 2 (u)
2t sin(2u) = --------
1 + t 2 or
2 tan(u) sin(2u) = -----------
1 + tan 2 (u)
2t tan(2u) = -------
1 - t 2 or
2 tan(u) tan(2u)= -------------
1 - tan 2 (u)
These three formulas are called the t-formulas or the Half-Angle Formulas
Application:
The equation of a line d is y = 3 x + 4 . u = the angle fom the x-axis to this line.
We know that tan(u) = 3.
Line d' is the reflection of the the line y = 3 in d.
The the angle fom the x-axis to the line d' is 2u.
The slope of line d' is tan(2u).
2t 2 tan(u) 2 . 3 tan(2u) = ------- = ------------ = ---------- = - 0.75
1 - t 2 1 - tan 2 (u) 1 - 9
We know that cos(u + v) = cos(u).cos(v)-sin(u).sin(v) cos(u - v) = cos(u).cos(v)+sin(u).sin(v) sin(u + v) = sin(u).cos(v)+cos(u).sin(v) sin(u - v) = sin(u).cos(v)-cos(u).sin(v) and from this, we have cos(u + v) + cos(u - v) = 2.cos(u).cos(v) cos(u + v) - cos(u - v) = -2.sin(u).sin(v)

sin(u + v) + sin(u - v) = 2. sin(u).cos(v) sin(u + v) - sin(u - v) = 2. cos(u).sin(v)
Let x = u + v and y = u - v then u = (1/2)(x + y) and v = (1/2)(x - y)
Now we have cos(x) + cos(y) = 2 cos((1/2)(x + y)) cos((1/2)(x - y)) cos(x) - cos(y) = -2 sin((1/2)(x + y)) sin((1/2)(x - y)) sin(x) + sin(y) = 2 sin((1/2)(x + y)) cos((1/2)(x - y)) sin(x) - sin(y) = 2 cos((1/2)(x + y)) sin((1/2)(x - y))
Simpson formulas
x + y x - y cos(x) + cos(y) = 2 cos ------ cos -------
2 2
x + y x - y cos(x) - cos(y) = -2 sin ------ sin -------
2 2
x + y x - y sin(x) + sin(y) = 2 sin ------ cos -------
2 2
x + y x - y sin(x) - sin(y) = 2 cos ------ sin -------
2 2
Example: The formulas can be used to factor trigonometric expressions.
cos(2x) - cos(2y)
-----------------
cos(2x) + cos(2y)
-2 sin(x+y) sin(x-y)
= --------------------
2 cos(x+y) cos(x-y)
= - tan(x+y) tan(x-y)
= tan(y+x) tan(y-x)
Many of the previous formulas can be used to factor trigonometric forms. This factoring can be done in many ways. With some examples, we'll show various methods for factorization.

 sin(2a).(1 + tan 2 (a))


 = 2 sin(a) cos(a) . (1/cos 2 (a))

 = 2 sin(a) / cos(a)

 = 2 tan(a)


 2 sin(2a) + sin(4a)

 = 2 sin(2a) + 2 sin(2a).cos(2a)

 = 2 sin(2a). (1 + cos(2a))

 = 4 sin(a) cos(a) .2 cos 2 (a)

 = 8 sin(a) cos 3 (a)


 1 - tan 4 (a)

 = (1 - tan 2 (a)).(1 + tan 2 (a))

 sin 2 (a) 1
 = ( 1 - ------- ) --------
 cos 2 (a) cos 2 (a)

 cos(2a)
 = --------
 cos 4 (a)


 cos 2 (a) - sin 2 (a) - 2 cos(a) + 1

 = cos 2 (a) - 2 cos(a) + cos 2 (a)

 = 2 cos 2 (a) - 2 cos(a)

 = 2 cos(a) (cos(a) - 1)

 = -2 cos(a) (1 - cos(a))

 = -2 cos(a) 2 sin 2 (a/2)

 = -4 cos(a) sin 2 (a/2)


 sin(2a) (1 + 2 cos(2a) ) + 2 sin(3a)

 = sin(2a) + 2 sin(2a) cos(2a) + 2 sin(3a)


 = sin(2a) + sin(4a) + 2 sin(3a)

 = 2 sin(3a) cos(a) + 2 sin(3a)

 = 2 sin(3a) (1+ cos(a))

 = 4 sin(3a) cos 2 (a/2)


 cos 2 (a) -2 cos(a) + cos(2a) + sin 2 (a)

 = 1 + cos(2a) - 2 cos(a)

 = 2 cos 2 (a) - 2 cos(a)

 = 2 cos(a) (cos(a) - 1)

 = -2 cos(a) (1 - cos(a))

 = -2 cos(a) 2 sin 2 (a/2)

 = -4 cos(a) sin 2 (a/2)


 (1 - cos(4a)) 2
 -----------------
 (1 - cos 2 (4a))


 (1 - cos(4a)) 2
 = ----------------------------
 (1 - cos(4a))(1 + cos(4a))

 (1 - cos(4a))
 = -------------------
 (1 + cos(4a))

 2 sin 2 (2a)
 = -----------------
 2 cos 2 (2a)

 = tan 2 (2a)


 cos 4 (a) - sin 4 (a)

 = (cos 2 (a) - sin 2 (a)) (cos 2 (a) + sin 2 (a))


 = (cos 2 (a) - sin 2 (a))

 = cos(2a)


 sin(a) + sin(b) + sin(c) - sin(a+b+c)

 = 2 sin( (a+b)/2 ) cos( (a-b)/2 ) + 2 cos( (a+b+2c)/2 ) sin( (-a-b)/2
)

 = 2 sin( (a+b)/2 ) ( cos( (a-b)/2 ) - cos( (a+b+2c)/2 )

 = -4 sin( (a+b)/2 ) sin( (a+c)/2 ) sin( (-b-c)/2 ) )

 = 4 sin( (a+b)/2 ) sin( (a+c)/2 ) sin( (b+c)/2 ) )


 sin 2 (a) - sin 2 (b) - sin 2 (a+b)

 = sin 2 (a) - sin 2 (b) - (sin(a) cos(b) + cos(a) sin(b)) 2

 = sin 2 (a) - sin 2 (b) - sin 2 (a) cos 2 (b) - cos 2 (a) sin 2 (b)
 - 2 sin(a) sin(b) cos(a) cos(b)

 = sin 2 (a) (1- cos 2 (b)) - sin 2 (b) (1 + cos 2 (a)) - 2 sin(a) sin(b) cos(a) cos(b)

 = sin 2 (a) sin 2 (b) - sin 2 (b) (1 + cos 2 (a)) - 2 sin(a) sin(b) cos(a) cos(b)

 = sin 2 (b) ( sin 2 (a) -1 - cos 2 (a)) - 2 sin(a) sin(b) cos(a) cos(b)

 = -2 sin 2 (b) cos 2 (a) - 2 sin(a) sin(b) cos(a) cos(b)

 = -2 sin(b) cos(a) ( sin(b) cos(a) sin(a) cos(b))

 = -2 cos(a) sin(b) sin(a+b)


 sin(2b+2c) (cos(2b) + cos(2c)) - sin(2b) - sin(2c)

 = sin(2b+2c) 2 cos(b+c) cos(b-c) - 2 sin(b+c)cos(b-c)

 = 4 sin(b+c) cos(b+c) cos(b+c) cos(b-c) - 2 sin(b+c)cos(b-c)

 = 2 sin(b+c) cos(b-c) ( 2 cos 2 (b+c) - 1)

 = 2 sin(b+c) cos(b-c) ( 2 cos 2 (b+c) - cos 2 (b+c) - sin 2 (b+c) )

 = 2 sin(b+c) cos(b-c) (cos 2 (b+c) - sin 2 (b+c) )


 = 2 sin(b+c) cos(b-c) cos(2b +2c)

We know that cos(u + v) = cos(u).cos(v)-sin(u).sin(v) cos(u - v) = cos(u).cos(v)+sin(u).sin(v) sin(u + v) = sin(u).cos(v)+cos(u).sin(v) sin(u - v) = sin(u).cos(v)-cos(u).sin(v)
Thus cos(u + v) + cos(u - v) = 2.cos(u).cos(v) cos(u + v) - cos(u - v) = -2.sin(u).sin(v) sin(u + v) + sin(u - v) = 2. sin(u).cos(v) sin(u + v) - sin(u - v) = 2. cos(u).sin(v) or
2.cos(u).cos(v) = cos(u + v) + cos(u - v)
-2.sin(u).sin(v) = cos(u + v) - cos(u - v)
2. sin(u).cos(v) = sin(u + v) + sin(u - v)
2. cos(u).sin(v) = sin(u + v) - sin(u - v)
The period of cos(u).cos(v) is equal to the period of cos(u + v) + cos(u - v)
The period of sin(u).sin(v) is equal to the period of cos(u + v) - cos(u - v)
The period of sin(u).cos(v) is equal to the period of sin(u + v) + sin(u - v)
The period of cos(u).sin(v) is equal to the period of sin(u + v) - sin(u - v)
Examples:
The period of cos(2x).sin(x+3) is equal to the period of sin(3x+3) - sin(x-3) and this period is 2 pi.
The period of cos(4x).cos(x/2) is equal to the period of cos(9x/2) + cos(7x/2) and this period is 4pi
The general sine function has y = a sin(b(x-c)) + d as equation, with a, b, c not negative and a and b not zero.
We can transform many equations of trigonometric functions to the form of a general sine function by using previous formulas.
We give some examples of such transformations.


 y = -3 sin(x)
 <=>
 y = 3 sin(x - pi)


 y = sin(-2x)
 <=>
 y = - sin(2x)
 <=>
 y = sin(2x - pi)
 <=>
 y = sin 2(x - pi/2)


 y = -4 sin(-3x)
 <=>
 y = 4 sin(3x)


 y = 2 cos(3x)
 <=>
 y = 2 sin(pi/2 - 3x)
 <=>
 y = -2 sin(3x -pi/2)
 <=>
 y = 2 sin(3x - 3pi/2)
 <=>
 y = 2 sin 3(x - pi/2)


 y = -2 cos(3x-1)
 <=>
 y = -2 sin(pi/2 -3x + 1)
 <=>
 y = 2 sin(3x - pi/2 -1)
 <=>
 y = 2 sin 3(x - (pi/6 + 1/3))


 y = cos(3x+4) - cos(3x-4)
 <=>
 y = -2 sin(3x) sin(4)
 <=>
 y = (-2sin(4)) sin(3x)


 y = sin(4x-3).cos(4x-3)
 <=>

 y = (1/2) sin(8x-6)
 <=>
 y = (1/2) sin(8(x-(3/4))

All the functions with an equation of the form y = a sin(u) + b cos(u) can be transformed to a general sine function.
This transformation is somewhat more difficult than in previous examples. a.sin(u)+b.cos(u) can be brought in the form A.sin(u-u o
).
Then, the transformation to the general sine function is easy.
a.sin(u) + b.cos(u)
= a( sin(u) + (b/a) cos(u) )
Take u o
such that tan(u o
) = - b/a
= a( sin(u) - tan(u o
) cos(u) )
= (a/cos(u o
)) . ( sin(u).cos(u o
) - sin(u o
).cos(u) )
Let A = (a/cos(u o
))
= A . sin(u - u o
)
Example
3 sin(x) - 2 cos(x)
= 3( sin(x) - (2/3) cos(x) )
Let tan(u o
) = 2/3 ; take u o
= 0.588
= 3( sin(x) - tan(u o
) cos(x) )
= (3/cos(u o
)) ( sin(x) cos(u o
) - cos(x) sin(u o
) )
= 3.6055 sin( x - 0.588)
cos(u) = cos(v)
With the help of the unit circle it is easy to see that
cos(u) = cos(v)

<=>
(u = v + k.2pi) or (u = -v + k.2pi) sin(u) = sin(v)
With the help of the unit circle it is easy to see that
sin(u) = sin(v)
<=>
(u = v + 2.k.pi) or (u = pi - v + 2.k.pi) tan(u) = tan(v)
With the help of the unit circle it is easy to see that
tan(u) = tan(v)
<=>
(u = v + k.pi) on condition that tan(u) and tan(v) exist cot(u) = cot(v)
With the help of the unit circle it is easy to see that
cot(u) = cot(v)
<=>
(u = v + k.pi) on condition that cot(u) and cot(v) exist
Example 1 cos(2x) = cos(pi-3x)
<=>
2x = (pi-3x) + 2.k.pi or 2x = -(pi-3x) + 2.k'.pi
<=>
5x = pi + 2.k.pi or -x = -pi + 2.k'.pi
<=> x = pi/5 + 2.k.pi/5 or x = pi - 2.k'.pi
Example 2 tan(x-pi/2) = tan(2x)
<=>
(x-pi/2) = 2x + k.pi
<=>
-x = pi/2 + k.pi
<=> x = -pi/2 - k.pi ( for these values tan(x-pi/2) and tan(2x)) exist)
Example 3

cos(x) = -1/3
<=> cos(x) = cos(1.91)
<=> x = 1.91 +2.k.pi or x = -1.91 - 2.k.pi
Example 4 sin(2x) = cos(x-pi/3)
<=> cos(pi/2 - 2x) = cos(x-pi/3)
<=> pi/2 - 2x = x - pi/3 + 2.k.pi or pi/2 - 2x = - x + pi/3 + 2.k'.pi
<=>
-3x = - pi/2 - pi/3 + 2.k.pi or -x = -pi/2 + pi/3 + 2.k'.pi
<=> x = pi/6 + pi/9 + 2.k.pi/3 or x = pi/2 - pi/3 - 2.k'.pi
<=> x = 5pi/18 + 2.k.pi/3 or x = pi/6 - 2.k'.pi
Example 5
3 sin(2x) = cos(2x)
<=>
3 tan(2x) = 1
<=> tan(2x) = 1/3
<=> tan(2x) = tan(0.32)
<=>
2x = 0.32 + k pi
<=> x = 0.16 + k pi/2
Example 6
tan(2x) . cot( x + pi/2) = 1
<=>
tan(2x) = tan( x + pi/2)
<=>
2x = x + pi/2 + k.pi
<=>
x = pi/2 + k.pi (on condition that tan(2x) and cot( x + pi/2) exist)
But cot( x + pi/2) does not exist for x = pi/2 + k.pi !!!!!
So, tan(2x) . cot( x + pi/2) = 1 has no solutions !
The expression " on condition that ...." is not redundant!
Example 1
2sin 2 (2x)+sin(2x)-1=0

<=> (let t = sin(2x) )
2t 2 + t - 1 = 0
<=> t = 0.5 or t = -1
<=> sin(2x) = 0.5 or sin(2x) = -1
<=> sin(2x) = sin(pi/6) or sin(2x) = sin(-pi/2)
<=>
2x = pi/6 +2.k.pi or 2x = pi - pi/6 +2.k.pi or
2x = -pi/2 +2.k.pi or 2x = pi + pi/2 +2.k.pi
<=> x = pi/12 + k.pi or x = 5pi/12 + k.pi or
x = -pi/4 + k.pi or x = 3pi/4 + k.pi
Sometimes it is convenient to view these solutions on the unit circle.
Example 2
cos 10x + 7 = 8 cos 5x
<=>
cos 10x - 8 cos 5x + 7 =0
<=>
1 + cos 10x - 8 cos 5x + 6 =0
<=>
2 cos 2 5x - 8 cos 5x + 6 =0
<=>
cos 2 5x - 4 cos 5x + 3 = 0
Let t = cos 5x
t 2 - 4t + 3 = 0
<=>
t = 3 or t = 1
<=>
cos 5x = 1
<=>
cos 5x = cos 0
<=>
5x = 2kpi
<=>
x = 2kpi / 5

Examples
In the same way, the following equations can be solved using an additional unknown. tan 2 (3x)+tan(3x)=0 sin 2 (x)(sin(x)+1)-0.25(sin(x)+1) = 0 cos(2x)+sin 2 (x) = 0.5 tan(2x)-cot(2x) = 1
Check your results by plotting graphs.
Example 1
3.sin(2x)-2.sin(x) = 0
<=>
6sin(x)cos(x)-2.sin(x) = 0
<=>
2.sin(x).(3cos()-1) = 0
<=> sin(x) = 0 or cos(x) = 1/3
<=> x = k.pi or x = 1.23 + 2.k.pi or x = -1.23 + 2.k'.pi
Examples
In the same way, the following equations can be solved using factoring. tan(x)tan(4x)+tan 2 (x) = 0 sin(7x)-sin(x) = sin(3x) cos(4x) + cos(2x) + cos(x) = 0 sin(5x)+sin(3x) = cos(2x)-cos(6x)
Check your results by plotting graphs.
First Method

First we'll show that a.sin(u)+b.cos(u) can be transformed in the form
A.sin(u-u o
) or in the form A.cos(u-u o
) .
a.sin(u) + b.cos(u)
= a( sin(u) + (b/a) cos(u) )
Take u o
such that tan(u o
) = - b/a
= a( sin(u) - tan(u o
) cos(u) )
= (a/cos(u o
)) . ( sin(u).cos(u o
) - sin(u o
).cos(u) )
Let A = (a/cos(u o
))
= A . sin(u - u o
)
= A . cos(pi/2 - u + u o
)
= A . cos(u - u o
')
Example
3 sin(x) - 2 cos(x)
= 3( sin(x) - (2/3) cos(x) )
Let tan(u o
) = 2/3 ; take u o
= 0.588
= 3( sin(x) - tan(u o
) cos(x) )
= (3/cos(u o
)) ( sin(x) cos(u o
) - cos(x) sin(u o
) )
= 3.6055 sin( x - 0.588)
or
= 3.6055 cos( x - 2.1598)
Plot the graph of 3 sin(x) - 2 cos(x) and the graph of 3.6055 sin( x - 0.588)
With this method we can solve the equation a.sin(u)+b.cos(u) = c
Example
3.sin(2x)+4.cos(2x) = 2
<=> sin(2x) + 4/3 .cos(2x) = 2/3
Let tan(t) = 4/3
<=> sin(2x) + tan(t) .cos(2x) = 2/3

<=> sin(2x)cos(t)+cos(2x)sin(t) = 2/3.cos(t)
<=> sin(2x+t) = 2/3.cos(t)
since 2/3.cos(t) = 0.4
<=> sin(2x+0.927) = sin(0.39)
<=>
2x + 0.927 = 0.39 +2.k.pi or 2x + 0.927 = pi - 0.39 +2.k'.pi
<=>
....
Second Method
Using the t-formulas
Example
3 sin(2x) + 4 cos(2x) = 2
Let tan(x) = t
<=>
2 t 1 - t 2
3 ------- + 4 -------- = 2
1 + t 2 1 + t 2
<=>
6 t + 4 - 4 t 2 = 2 + 2 t 2
<=>
6 t 2 - 6 t - 2 = 0
<=>
3 t 2 - 3 t -1 = 0
<=>
t = 1.26 or t = -0.26
<=>
tan(x) = 1.26 or tan(x) = -0.26
<=>
x = 0.9 + k pi or x = -0.26 + k pi

An equation is homogeneous in a and b if and only if we obtain an equivalent equation when we replace a and b by ra and rb (r is not 0). Example: a
3
x
2
+5 a.b
2
x +3 a
2
.b = 0 is an equation in x which is homogeneous in a en b.
Now, we have in view the equations which are homogeneous in sin(u) and cos(u).
Procedure
1.
Reduce the equation to the form F = 0. If possible, use factoring to the left hand side and solve the simple parts.
2.
Divide the remaining equation through by a suitable power of cos(u), such that tan(u) appears everywhere.
3.
Let t = tan(u) and solve the algebraic equation.
4.
Return to tan(u)
Example
2.cos
3 (x)+2.sin
2 (x)cos(x) = 5.sin(x)cos 2 (x)
<=> cos(x).(2.cos
2 (x)+2.sin
2 (x) - 5.sin(x)cos(x)) = 0
<=>
The simple part cos(x) = 0 gives us x = pi/2 + k.pi
In the second part, we divide both sides by cos 2 (x). Then we have
2.tan
2 (x) - 5.tan(x) +2 = 0
Let t = tan(x)
<=>
2.t
2 - 5 t + 2 = 0
<=> t = 0.5 or t = 2
<=> tan(x) = 0.5 or tan(x) = 2
<=> x = 0.464 +k.pi or x = 1.107 +k.pi

Some equations can be solved, in an appropriate manner, by combining various discussed methods. In addition, some trigonometric formulas are often used to transform the equation into a suitable form. Moreover, experience and understanding play an important role by solving difficult equations. We give a not obvious example.
1/sin(x) + 1/cos(x) = sqrt(3)
1/sin(x) + 1/cos(x) = sqrt(3)
sin(x) + cos(x)
<=> ------------------- = sqrt(3)
sin(x) cos(x)
<=> sin(x) + cos(x) = sqrt(3) sin(x) cos(x) (1)
--------------------------------------------------------
If we square both sides of the equation (1), only the product sin(x)cos(x) occurs. Then, we can find the value of sin(x)cos(x) and with this we'll simplify the equation (1).
From (1) it follows
( sin(x) + cos(x)) 2 = 3 (sin(x) cos(x)) 2
<=> 1 + 2 sin(x) cos(x) = 3 (sin(x) cos(x)) 2
Let y = sin(x) cos(x)
<=> 3 y 2 - 2 y -1 = 0
<=> y = 1 or -1/3
<=> sin(x) cos(x) = 1 or sin(x) cos(x) = -1/3
If sin(x) cos(x) = 1 then sin(2x) = 2 and this is impossible.
Conclusion: From (1) it follows that
sin(x) cos(x) = -1/3 (2)
----------------------------------------------------------
Now, we use this result. We bring (2) in (1)
sin(x) + cos(x) = - sqrt(3)/3 (3)
If (1) is true, then (2) is true and thereby (3) is true.
Now, we show that the reverse is true. We start with (3).
sin(x) + cos(x) = - sqrt(3)/3

=> (sin(x) + cos(x) 2 = 1/3
=> 1 + 2 sin(x) cos(x) = 1/3
=> sin(x) cos(x) = -1/3 and this is (2)
So, If (3) is true , then (2) is true and thereby (1) is true.
Conclusion:
(1) and (3) are equivalent equations. We'll solve (3) now.
-------------------------------------------------------------
sin(x) + cos(x) = - sqrt(3)/3
<=> cos(pi/2 -x) + cos(x) = - sqrt(3)/3
<=> 2 cos(pi/4) cos (pi/4-x) = - sqrt(3)/3
<=> sqrt(2) cos (pi/4-x) = - sqrt(3)/3
<=> cos (pi/4-x) = -1/ sqrt(6)
Let a = arccos(-1/sqrt(6))
<=> cos (pi/4-x) = cos(a)
<=> pi/4 - x = a + 2 k pi or pi/4 - x = -a + 2 k pi
<=> x = pi/4 - a + 2 k pi or x = pi/4 + a + 2 k pi
The solutions of the given equation are the values pi/4 - a + 2 k pi en pi/4 + a + 2 k pi with a = arccos(-1/sqrt(6))
A second method
Many equations can be solved in different ways. We'll show an alternative way to solve the equation
1/sin(x) + 1/cos(x) = sqrt(3)
The period of the function on the left-hand side is 2 pi. If we have the solutions to the equation in
[0, 2pi], then we know all solutions.
First, we calculate the solutions in [0, 2pi]. Since the right-hand side of the equation is positive, solutions are only possible when the left-hand side is positive too.
By plotting the function 1/sin(x) + 1/cos(x), we see that the image is positive in the intervals
(0,pi/2) ; (3pi/4, pi) and (3pi/2, 7pi/4).

The solutions can only occur in these intervals. If we restrict the values of x to these intervals, then both sides of the equation are positive and we can write
1/sin(x) + 1/cos(x) = sqrt(3)
<=> (1/sin(x) + 1/cos(x)) 2 = 3
sin(x) + cos(x)
<=> (------------------) 2 = 3
sin(x) cos(x)
<=> 1 + 2 sin(x) cos(x) = 3 sin 2 (x) cos 2 (x)
Let y = sin(x) cos(x)
<=> 3 y 2 - 2 y - 1 = 0
<=> y = 1 or y = -1/3
Case y = 1
sin(x) cos(x) = 1
<=> 2 sin(x) cos(x) = 2
<=> sin(2x) = 2
In this case, there are no solutions
Case y = -1/3
2 sin(x) cos(x) = -2/3
<=> sin(2x) = -2/3
let b = arcsin(-2/3) ; b = -0.7297
<=> sin(2x) = sin(b)
<=> 2x = b + 2 k pi or 2x = (pi-b) + 2kpi
<=> x = b/2 + k pi or x = pi/2 - b/2 + k pi
Now, we will take only the x-values located in the intervals
(0,pi/2) ; (3pi/4, pi) en (3pi/2, 7pi/4).
There are 2 solutions :
x = b/2 + pi = 2.7767 and
x = pi/2 - b/2 + pi = 3pi/2 - b/2 = 5.077
All solutions to the equation 1/sin(x) + 1/cos(x) = sqrt(3) are

b/2 + pi + 2 k pi en 3pi/2 - b/2 + 2 k pi
k is an integer.
'=<' means equal or less than
'>=' means equal or greater than
 sin(x/2) > 1/2
We draw the solutions for (x/2) on the unit circle.
Now, we see that:
sin(x/2) > 1/2
<=> pi/6 + 2 k pi < x/2 < 5pi/6 + 2 k pi
<=> pi/3 +4 k pi < x < 5pi/3 + 4 k pi
For each k, we have an open interval with solutions.
The solution set V is the union of all these open intervals.

V = { U (pi/3 +4 k pi , 5 pi/3 + 4 k pi) | k in Z }
k
 tan(2x) < 1/3
We draw the solutions for (2x) on the unit circle.
Now, we see that:
tan(2x) < 1/3
<=> -pi/2 + k pi < 2x < 0.32 + k pi
<=> -pi/4 + k pi/2 < x < 0.16 + k pi/2
For each k, we have an open interval with solutions.
The solution set V is the union of all these open intervals.
V = { U (-pi/4 + k pi/2 , 0.16 + k pi/2) | k in Z }
k
 tan(2x + pi/5 ) < 1/3
This is a variation on the previous example.
The figure is the same as the previous one, but now it gives the solutions for (2x + pi/5).
Now we have :
tan(2x + pi/5 ) < 1/3

<=> -pi/2 + k pi < 2x + pi/5 < 0.32 + k pi
<=> -pi/2 -pi/5 + k pi < 2x < 0.32 - pi/5 + k pi
<=> -7 pi/20 + k pi/2 < x < 0.16 -pi/10 + k pi/2
The solution set V is:
V = { U (-7 pi/20 + k pi/2 , 0.16 - pi/10 + k pi/2) | k in Z }
k
 2 sin
2
(x) - 3 sin(x) + 1 = < 0
Let t = sin(x) .
2 t 2 - 3 t + 1 < 0
<=> 2 (t - 1)(t - 1/2) < 0
A sign study of the left hand side gives
<=> 1/2 =< t =< 1
<=> 1/2 =< sin(x) =< 1
Draw the solutions for x on the unit circle. We see that:
<=> pi/6 + 2 k pi =< x =< 5pi/6 + 2 k pi
The solution set is
V = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi] | k in Z }
k
 Another method to solve 2 sin
2
(x) - 3 sin(x) + 1 = < 0
One can also factor the left hand side directly and investigate the sign in a period-interval.
2 sin 2 (x) - 3 sin(x) + 1
<=> 2 (sin(x) - 1)(sin(x) - 1/2)
We take a simple period-interval [0,2pi). We investigate the sign of each factor.

x 0 pi/6 pi/2 5pi/6 pi 2pi
---------------------------------------------------------------
sin(x)-1 - - - - - - 0 - - - - - - - - - - - - - -
--------------------------------------------------------------- sin(x)-1/2 - - 0 + + + + + + 0 - - - - - - - - - -
---------------------------------------------------------------
product + + 0 - - - 0 - - 0 + + + + + + + + + +
---------------------------------------------------------------
The solutions in the period-interval are pi/6 =< x =< 5pi/6
The solution set is :
V = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi] | k in Z }
k
 cosec(x) < sec(x)
We investigate first the inequality in the period-interval [0,2pi).
The values 0 ; pi/2 ; pi ; 3pi/2 are no solutions . We investigate the other values of x in each quadrant. o First quadrant o o cosec(x) < sec(x) o o <=> 1/ sin(x) < 1 / cos(x) o o now we have sin(x).cos(x) > 0 o o <=> cos(x) < sin(x)
The solution set is ( pi/4, pi/2 ). o Second quadrant
Now we have cos(x) < 0 and sin(x) > 0. There are no solutions. o Third quadrant o o cosec(x) < sec(x) o o <=> 1/ sin(x) < 1 / cos(x) o o now we have sin(x).cos(x) > 0 o o <=> cos(x) < sin(x)

The solution set is ( pi , 5 pi/4 ) o Fourth quadrant Now we have cos(x) > 0 and sin(x) < 0.
The solution set is ( 3 pi/2 , 2pi)
The solution set of the given inequality is the union of all the intervals
(pi/4 + 2 k pi , pi/2 + 2 k pi )
(pi + 2 k pi , 5pi/4 + 2 k pi )
(3pi/2 + 2 k pi , 2 pi + 2 k pi ) with k in Z
 First transform, then solve.

 cot(x) + 1
 ------------- > 0
 sin (x)


 cot(x) + 1
 <=> ------------- > 0 and sin (x) not 0
 sin (x)


 cos(x) + sin(x)
 <=> ------------------- > 0 and sin (x) not 0
 sin 2 (x)


 <=> cos(x) + sin(x) > 0 and sin (x) not 0


 <=> sin(x) + sin(pi/2 -x) > 0 and sin (x) not 0

 with Simpson's formulas

 <=> 2 sin(pi/4) cos( x - pi/4) > 0

 <=> cos( x - pi/4) > 0 and sin (x) not 0
Using the unit circle we see that
<=> -pi/2 + 2k pi < x - pi/4 < pi/2 + 2k pi and sin (x) not 0
<=> -pi/4 + 2k pi < x < 3pi/4 + 2k pi and sin (x) not 0

The solution set V is the union of the open intervals
V = { U (-pi/4 + 2k pi , 3pi/4 + 2k pi ) | k in Z } \ { k pi | k in Z }
k
All equations are solved using the same method. We replace the equation successively by a necessary condition . This means that the values that we find are not necessary solutions.
Afterwards, we have to test these values against the initial equation. The false or parasitic values must be deleted.
Example 1
arcsin(2x) = pi/4 + arcsin(x)
<=>
/ arcsin(2x) = a
| arcsin(x) = b (1)
\ a = pi/4 + b
/ sin(a) = 2x
=> | sin(b) = x
\ a = pi/4 + b
=> / sin(pi/4 + b) = 2x
\ sin(b) = x
sum formulas
=> / cos(b) + sin(b) = 2x.sqrt(2)
\ sin(b) = x
=> / cos(b) = 2x.sqrt(2) - x
\ sin(b) = x
=> (2x.sqrt(2) - x) 2 + x 2 = 1
=> ....
=> x = +0.4798 or x = -0.4798
We test these values against the initial equation. The only solution is 0.4798.
The other x-value is false or parasitic.
Example 2

arctan(x+1) = 3.arctan(x-1)
<=>
/ arctan(x+1) = a
| arctan(x-1) = b
\ a = 3 b
=> / tan(a) = x + 1
| tan(b) = x - 1
\ a = 3 b
=> / tan(3b) = x+ 1
\ tan(b) = x - 1
3 tan(b) - tan 3 (b)
but tan(3b) = --------------------
1 - 3.tan
2 (b)
3(x-1) - (x-1) 3
=> x+1 = --------------------
1 - 3 (x-1) 2
=> (x+1) (1 - 3 (x-1) 2 ) = 3(x-1) - (x-1) 3
=> ...
=> x = 0 or x = sqrt(2) or x = -sqrt(2)
We test these values against the initial equation. The only solution is sqrt(2).
The other x-values are false or parasitic.
Example 3
arctan(x) + arctan(2x) = pi/4
<=>
/ arctan(x) = a
| arctan(2x) = b
\ a + b = pi/4
=> / x = tan(a)
| 2x = tan(b)
\ a + b = pi/4
=> / x = tan(a)
\ 2x = tan(pi/4-a)
1 - tan(a)
but tan(pi/4-a) = ---------------- since tan(pi/4) = 1
1 + tan(a)

1 - x
=> 2x = ----------
1 + x
=> ...
=> x = (-3+sqrt(17))/4 or x = (-3-sqrt(17))/4
We test these values against the initial equation. The only solution is (-3+sqrt(17))/4.
The other x-value is false or parasitic.
Example 4
arctan( (x+1)/(x+2) ) - arctan ( (x-1)/(x-2) ) = arccos( 3/sqrt(13) )
/ arctan( (x+1)/(x+2) ) = a
<=> | arctan( (x-1)/(x-2) ) = b
| arccos( 3/sqrt(13) ) = c
\ a - b = c
/ tan(a) = (x+1)/(x+2)
=> | tan(b) = (x-1)/(x-2)
| cos(c) = 3/sqrt(13)
\ a - b = c
tan(a) - tan(b)
but tan(a-b) = ------------------ and after calculation one finds
1 + tan(a) tan(b)
-2 x
/ tan(c) = ----------------
=> | 2 x 2 - 5
|
\ cos(c) = 3/sqrt(13)
From the last equation it follows
1 + tan 2 (c) = 1/cos 2 (c) = 13/9 => tan 2 (c) = 4/9
There are now two cases
First case :
-2 x
/ tan(c) = ----------------
=> | 2 x 2 - 5
|
\ tan(c) = 2/3
=> ....

=> x = 1 or x = -5/2
Second case:
-2 x
/ tan(c) = ----------------
=> | 2 x 2 - 5
|
\ tan(c) = - 2/3
=> ....
=> x = -1 or x = 5/2
We test these values against the initial equation. The only solutions are 1 and -5/2.
The other x-values are false or parasitic.
In the following examples, 'less or equal' is written as ' =< '.
Example 1
________
| 2
\| 1 - p
Show that cot(arcsin(p)) = -----------
p
let b = arcsin(p) ,then sin(b) = p with b in [-pi/2 , pi/2].
So, cos(b) = sqrt( 1 - p 2 ) and
________
| 2
\| 1 - p
cot(arcsin(p)) = cot(b) = ---------
p
Example 2
Prove that the following equation has no solutions for x > 0
cos(arctan(x)) + x sin(arctan(x)) = x

Let u = arctan(x) ; Since x > 0 is u in (0, pi/2) and tan(u) = x
1 + tan 2 (u) = 1 + x 2
<=>
1/ cos 2 (u) = 1 + x 2
<=>
cos(u) = 1/ sqrt(1+x 2 )
- - - - - - - - - - - - - - - -
sin(u) = tan(u) . cos(u)
<=> sin(u) = x / sqrt(1+x 2 )
- - - - - - - - - - - - - - - -
cos(arctan(x)) + x sin(arctan(x)) = x
<=>
cos(u) + x sin(u) = x
<=>
1/ sqrt(1+x 2 ) + x 2 / sqrt(1+x 2 ) = x
<=>
1 + x 2 = x . sqrt(1+x 2 )
<=>
(1 + x 2 ) 2 = x 2 . (1+x 2 )
<=>
1 + x 2 = x 2
and this equation has no solutions.
Example 3
Find the domain of arccos(arcsin(x))
x belongs to the domain of arccos(arcsin(x))
<=>
- 1 =< arcsin(x) =< 1
and since the sine function increases in [-1,1]
<=>
sin(-1) =< x =< sin(1)
Conclusion : The domain of arccos(arcsin(x)) is [ -sin(1), sin(1) ]
Example 4

Find the domain of arcsin(arccos(x))
x belongs to the domain of arcsin(arccos(x))
<=>
- 1 =< arccos(x) =< 1
<=>
0 =< arccos(x) =< 1
and since the cosine function decreases in [0,1]
<=>
cos(0) >= x >= cos(1)
Conclusion : The domain of arcsin(arccos(x)) is [cos(1),1]
Example 5 f(x) = arccos(a. arcsin(x)) with a > 0
Find the values of a such that the domain D of f(x) is as large as possible.
Calculate the value of a such that the domain D = [-0.5 ; 0.5]
x belongs to the domain of arccos(a. arcsin(x))
<=>
- 1 =< a . arcsin(x) =< 1
<=>
-1/a =< arcsin(x) =< 1/a (*)
First case: 1/a =< pi/2
Now (*) is equivalent with
sin(-1/a) =< x =< sin(1/a) (**)
In this case the domain D is maximum if and only if sin(1/a)is maximum.
Then a = 2/pi.
Second case: 1/a > pi/2
Now (*) is equivalent with
-pi/2 =< arcsin(x) =< pi/2
In this case the domain D is always [-1,1].

Conclusion: If a =< 2/pi , then the domain D is as large as possible.
From (**) it follows that
domain D is [-1/2,1/2]
<=>
sin(1/a) = 1/2
<=>
a = 6/pi
Remark : you can illustrate and explore many of the previous steps with a function plotter.
Example 6
Prove that the following expression holds for all the positive integers n.
1
arctan -------------- = arctan(1/n) - arctan(1/(n+1))
n 2 + n + 1
We note first that
1 1 1
0 < ------------- < -------- < --- =< 1
n 2 + n + 1 n+1 n and therefore
1 1 1
0 < arctan ------------- < arctan -------- < arctan --- =< pi/4
n 2 + n + 1 n+1 n
This means that the expressions
1
arctan ------------- and arctan(1/n) - arctan(1/(n+1))
n 2 + n + 1 are in (0, pi/4 ] for all n.
Therefore, the expression
1
arctan -------------- = arctan(1/n) - arctan(1/(n+1))
n 2 + n + 1 is equivalent with
1
------------- = tan ( arctan(1/n) - arctan(1/(n+1)) )
n 2 + n + 1

We simplify the right side
tan ( arctan(1/n) - arctan(1/(n+1)) )
tan (arctan(1/n)) - tan (arctan(1/(n+1)))
= ----------------------------------------------
1 + tan (arctan(1/n)) . tan (arctan(1/(n+1)))
1/n - 1/(n+1)
= -------------------------
1 + (1/n)(1/(n+1))
1
= -------------
n 2 + n + 1
Remark : you can illustrate and explore many of the previous steps with a function plotter.
Click here to find the basic formulas.
You can find solved problems about trigonometry using this link.
 trigbook (pdf)
Triangles ; identities ; graphs ; inverse functions ; equations ; polar coordinates
 triangles (pdf)
Many examples and exercises about triangles
 Formulas sheet (pdf)

4 pages Trig formulas
Trigonometry (pdf)
 ratios ; functions ; identities ; equations ; examples ; solved problems
SOS-Math Trigonometry
Topics and Problems
MATH-abundance home page - tutorial

MATH-tutorial Index
The tutorial address is http://home.scarlet.be/math/
Copying Conditions
Send all suggestions, remarks and reports on errors to Johan.Claeys@ping.be The subject of the mail must contain the flemish word 'wiskunde' because other mails are filtered to Trash